Question 4 of 6: Question 4 (paper Question IV): Rod released against a compressed spring
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3 (Statics and Dynamics) — 2016-May. 3 hours, closed book (one 8.5″×11″ formula sheet permitted). Part A (Statics) offers 3 questions, answer any 2; Part B (Dynamics) offers 3 questions, answer any 2. Every printed question is solved below (all 6), even though the exam instructions require only 4.
Given. Slender rod AB, mass 15 kg, length 2 m, pivoted at O (0.4 m from B, so 1.6 m from A); mass centre C is the midpoint (1 m from each end), i.e. 0.6 m from O on the A-side. Spring $k=300\text{ kN/m}$ compressed $x=25\text{ mm}$ at end A; rod starts horizontal and is released from rest.
Given data
Quantity
Value
Mass, m
15 kg
Length, L
2 m
Pivot O to B
0.4 m
Pivot O to mass centre C
0.6 m
Spring constant, k
300 kN/m
Spring compression, x
25 mm
Find. Angular velocity $\omega$ and pivot reaction $\mathbf{O}$ as the rod passes through the vertical position (end A above O).
Fig. 4 – Position 1 (horizontal, spring compressed at A) and Position 2 (vertical, A above O), with the reaction at O.
Approach. The spring releases its stored energy over the first 25 mm of travel (essentially at the horizontal position), so a single work-energy equation from Position 1 (horizontal, at rest, spring compressed) to Position 2 (vertical) gives $\omega$. At the exact vertical position the mass centre C lies on the same vertical line through O as the weight force, so gravity's moment about O – and hence the angular acceleration $\alpha$ – is momentarily zero; Newton's second law for the mass centre (pure centripetal acceleration) then gives the pivot reaction.
Mass moment of inertia about O. $I_C=\tfrac{1}{12}mL^2=\tfrac{1}{12}(15)(2)^2=5.00\text{ kg}\cdot\text{m}^2$; by the parallel-axis theorem with $d_{OC}=0.6\text{ m}$:
$$I_O=I_C+md_{OC}^2=5.00+15(0.6)^2=10.40\text{ kg}\cdot\text{m}^2$$
Energy stored in the spring.
$$V_{\text{spring}}=\tfrac12kx^2=\tfrac12(300{,}000)(0.025)^2=93.75\text{ J}$$
Rise in gravitational PE of C. Going from horizontal (C at O's height) to vertical with A above O, C rises by $d_{OC}=0.6\text{ m}$:
$$\Delta V_{\text{grav}}=mg\,d_{OC}=15(9.81)(0.6)=88.29\text{ J}$$
Conservation of energy, $T_1+V_1=T_2+V_2$ with $T_1=0$:
$$\tfrac12I_O\omega^2=93.75-88.29=5.46\text{ J}\ \Rightarrow\ \boxed{\omega=1.025\text{ rad/s}}$$
Angular acceleration at the vertical instant. With C directly above O, the weight's moment arm about O is zero, so $\Sigma M_O=I_O\alpha=0\Rightarrow\alpha=0$; the mass centre therefore has only centripetal acceleration, $a_G=\omega^2d_{OC}$ directed from C toward O (downward):
$$a_G=(1.025)^2(0.6)=0.630\text{ m/s}^2\ \text{(down)}$$
Newton's second law for the rod. $\Sigma F_x=ma_{Gx}=0\Rightarrow O_x=0$. $\Sigma F_y=ma_{Gy}$: with the reaction $O_y$ up and weight down,
$$O_y-mg=-m\omega^2d_{OC}\ \Rightarrow\ \boxed{O_y=m(g-\omega^2d_{OC})=15(9.81-0.630)=137.7\text{ N}}$$
